 Solution

Suppose $f$ is sufficiently smooth and equivariant, $f(0,\mu )=0$, and the group representation on the critical space is absolutely irreducible. Its linearization is then $D_{x}f(0,\mu )=a(\mu )I$. Assume $a(0)=0$ and $a^{\prime }(0)\ne 0$. The equivariant branching lemma states that each isotropy subgroup $H$ with $\dim \mathrm{Fix}(H)=1$ has a local nontrivial branch of equilibria with that isotropy, apart from the common origin. Conjugate isotropy subgroups give symmetry-related branches, so branch types are counted up to conjugacy.

To see the scalar mechanism, write the fixed-point subspace of a group action as $\mathbb{R}e$ and restrict $f(se,\mu )=eh(s,\mu )$. Equivariance makes the fixed-point subspace of a group action invariant. Since $h(0,\mu )=0$, write $h=sk(s,\mu )$; $k(0,0)=0$ and $k_{\mu }(0,0)=a^{\prime }(0)\ne 0$. The implicit function theorem solves $k(s,\mu (s))=0$ near zero. For small nonzero $s$, the stabilizer subgroup is exactly that of $e$. A normalizer element reversing $e$ makes $h$ odd and $\mu (s)$ even; the generic branch is then a pitchfork bifurcation. The lemma ensures existence, not dynamical stability, and does not exclude further branches of smaller isotropy.